# Thread: integration problem

1. ## integration problem

Hi guys,

I'm super stuck with these two problems on one of my practice exams, can anyone help me out?

Find the integral between 4 and 3 of (u^2 + 1) / (u - 2)^2
and the antiderivative of cosxe^(sinx) dx

It asks to find the volume of the solid of revolution obtained by rotating, a full turn about the x-axis, the area between the x-axis and the curve y = 2 sin x, for x ∈ [ π , 3π ].

Thnanks

2. Hi
Originally Posted by katie
I'm super stuck with these two problems on one of my practice exams, can anyone help me out?

Find the integral between 4 and 3 of (u^2 + 1) / (u - 2)^2
The hint is to rewrite the integral as follows (using $(u-2)=u^2-4u+4$ )


\begin{aligned}
\int_3^4\frac{u^2+1}{(u-2)^2}\,\mathrm{d}u &= \int_3^4\frac{u^2-4u+4+4u-4+1}{(u-2)^2}\,\mathrm{d}u\\
&=\int_3^4\frac{u^2-4u+4}{(u-2)^2}\,\mathrm{d}u+ \int_3^4\frac{4u-3}{(u-2)^2}\,\mathrm{d}u\\
&=\int_3^4\mathrm{d}u+ \int_3^4\frac{4u-3}{(u-2)^2}\,\mathrm{d}u\\
\end{aligned}

For the last integral, let's make appear the derivative of the denominator ( $2u-4$) at the numerator so that one can easily find an anti-derivative :

\begin{aligned}
\int_3^4\frac{4u-3}{(u-2)^2}\,\mathrm{d}u&=2\int_3^4\frac{2u-\frac{3}{2}}{(u-2)^2}\,\mathrm{d}u\\
&=2\int_3^4\frac{2u-4+4-\frac{3}{2}}{(u-2)^2}\,\mathrm{d}u\\
&=2\int_3^4\frac{2u-4}{(u-2)^2}\,\mathrm{d}u+2\int_3^4\frac{4-\frac{3}{2}}{(u-2)^2}\,\mathrm{d}u\\
&=2\int_3^4\frac{2u-4}{(u-2)^2}\,\mathrm{d}u+5\int_3^4\frac{1}{(u-2)^2}\,\mathrm{d}u\\
\end{aligned}

If you don't see the what can be done with the last two integrals, substitute $x=u-2$.

Hope that helps.

3. Originally Posted by katie
Hi guys,

I'm super stuck with these two problems on one of my practice exams, can anyone help me out?

Find the integral between 4 and 3 of (u^2 + 1) / (u - 2)^2

Mr F says: ${\color{red} \frac{u^2 + 1}{u^2 - 4u + 4} = \frac{(u^2 - 4u + 4) + (4u - 3)}{u^2 - 4u + 4} = 1 + \frac{4u - 3}{u^2 - 4u + 4}}$.

Now note that ${\color{red} \frac{4u - 3}{u^2 - 4u + 4} = \frac{(4u - 8) + 5}{u^2 - 4u + 4} = \frac{4u - 8}{u^2 - 4u + 4} + \frac{5}{u^2 - 4u + 4}}$.

Therefore ${\color{red} \frac{u^2 + 1}{(u-2)^2} = 1 + \frac{5}{(u-2)^2} + \frac{4u - 8}{u^2 - 4u + 4}}$.

See what you can do with that.

and the antiderivative of cosxe^(sinx) dx

Mr F says: Make the substitution ${\color{red} u = \sin x}$.

It asks to find the volume of the solid of revolution obtained by rotating, a full turn about the x-axis, the area between the x-axis and the curve y = 2 sin x, for x ∈ [ π , 3π ].

Mr F says: By symmetry, it'll be twice the volume when the area from 0 to pi is rotated:

${\color{red} V = 2 \pi \int_{0}^{\pi} (2 \sin x)^2 \, dx = ......}$.

Thnanks
..